I'm not quite sure which region should be shaded. Is there anyone who can confirm the if the answer is correct?
Find the range of values of x which satisfies this inequality?
(2x+1)(5-3x) < 0
I calculated the inequality as this is my answer
x < -1/2 and x < 5/3
-1/2 < x < 5/3 which means the region between -1/2 and 5/3 should be shaded ?
On
From your inequality you immediately see that the zeros of the function $f(x) = (2x+1)(5-3x) = -6x^2+7x + 5$ are $$x_1 = -\frac{1}{2} \quad \text{and} \quad x_2= \frac{3}{5}.$$ The function $f$ is a quadratic function. In your case it's a convex parabola (headed downwards). The function hence has to take negative values for $x$ outside the interval $\left[-\frac{1}{2}, \frac{3}{5}\right]$. Therefore the solutions $x$ of your inequality belong to the set $$\Bbb{L}=]-\infty, -\frac12 [ \,\, \cup \,\, ]\frac35, + \infty[.$$
You should have $x > \frac{5}{3}$ instead (try it with $x=2$ for example), so the region to be shaded is actually $x < -\frac{1}{2}$ and $x > \frac{5}{3}$.
In general, once you have the factored form, you can make a sign diagram to find the shaded region:
$$ \begin{array}{c|lcr} \text{sign of $f(x)$} & - & + & - \\ \hline x & -\frac{1}{2} & 0 & \frac{5}{3} \\ \end{array} $$
You can also notice that the parabola has a negative $x^2$ coefficient, so the parabola is concave down, from which you can get to the same conclusion.
Using a graphing tool such as Desmos or Wolfram Alpha will give you the answer if you are really not sure, but not how to find the answer (unless you have WA Pro).